You can rewrite the equation as $x = y^{xy^{-2}}$, plugging this into itself gives $x = y^{y^{xy^{-2}-2}}$, and by taking logs (and supposing $y > 1$) we get $y^{xy^{-2}-2} = xy^{-2}$. Finally, raising to the $y^2$th power, we get the equality
$y^{x-2y^2} = (xy^{-2})^{y^2}$

If $x > 2y^2$, the LHS is an integer, the RHS is a rational to an integer exponent.
This forces the rational on the right to be an integer, and so $x$ has to be a multiple of $y^2$. So $x = my^2$ for some $m > 2$. Then the equation simplifies to $y = m^{1/(m-2)}$ which has the two solutions for $m=3$ and $m=4$

If $x \le 2y^2$ we can take the inverses and now it is $y^2$ that has to be a multiple of $x$.
If $y^2 = mx$, we get $(x^{2m})^x = (mx)^x$ thus $x = m^{1/(2m-1)} < 2$, and this is possible only for $x=y=m=1$.